Uses of Implicit differentiation

Question

I would like to find $\frac{\mathrm dy}{\mathrm dx}$, where we have the implicit equation $$(2x+y)^4+3x^2y=y^3.$$ I just can't seem to understand the problem and which rule of calculus to use. Could someone please further elaborate?

Answer 1

Just use the chain rule, the power rule, and the product rule and differentiate both sides of the equation.

$4(2x+y)^3(2+y')+3x^2y'+6xy=3y^2y'$

$y'(4(2x+y)^3+3x^2-3y^2)=-6xy-8(2x+y)^3$

$y'=\frac{-6xy-8(2x+y)^3}{4(2x+y)^3+3x^2-3y^2}$

Answer 2

If we differentiate the identity we get

$$2\cdot dx+dy+6xy\cdot dx+3x^2\cdot dy=3y^2\cdot dy$$

Rearranging we get

$$\left(2+6xy\right)\cdot dx=\left(3y^2-3x^2-1\right)\cdot dy$$

And we deduce

$${dy\over dx}={2+6xy\over 3y^2-3x^2-1}$$

This is valid of course outside the hyperbola defined by $x^2-y^2=-{1\over 3}$

Answer 3

Notice, $$(2x+y)+3x^2y=y^3$$ differentiating both the sides w.r.t. $x$ as follows $$\frac{d}{dx}(2x+y+3x^2y)=\frac{d}{dx}(y^3)$$ $$\frac{d}{dx}(2x)+\frac{d}{dx}(y)+\frac{d}{dx}(3x^2y)=\frac{d}{dx}(y^3)$$ $$2(1)+\frac{dy}{dx}+3x^2\frac{d}{dx}(y)+3y\frac{d}{dx}(x^2)=3y^2\frac{dy}{dx}$$

$$2+\frac{dy}{dx}+3x^2\frac{dy}{dx}+6xy=3y^2\frac{dy}{dx}$$ $$\frac{dy}{dx}(3x^2-3y^2+1)=-(6xy+2)$$

$$\frac{dy}{dx}=\color{red}{\frac{-(6xy+2)}{3x^2-3y^2+1}}$$